Question

find the zeros of y=x^3-x^2-9x+9
with explanation

Answer 1

Answer:

So, we have the equation:

$y=x^3-x^2-9x+9$

Usually, we have to do trial-by-error methods like using the Rational Root Theorem to factor cubic polynomials. However, while this may seem like a complicated cubic equation to factor, it has a simple trick to make it much easier!

We can actually factor this polynomial through Factoring by Grouping. With this method, we can factor a cubic polynomial. We split the equation into two parts, like this:

$y=x^3-x^2-9x+9$

$y=(x^3-x^2)+(-9x+9)$

Which can be simplified to:

$y=(x^2)(x-1)+(-9)(x-1)$

We took out a factor of x^2 for the first term and a negative 9 for the second term. As we can see, we now have two terms that are the same:

(x - 1)

We can factor out the (x - 1) like this:

$y=(x^2)(x-1)+(-9)(x-1)$

$y=(x-1)(x^2-9)$

Now, this is not the most simplified form. See the x^2-9 term? We can use the Difference of Squares.

In general:

$a^2-b^2=(a-b)(a+b)$

We have the expression:

$(x^2-9)=(x^2-3^2)$

$(x^2-3^2)=(x-3)(x+3)$

Therefore:

$y=(x-1)(x^2-9)$

$y=(x-1)(x-3)(x+3)$

Our fully factored form for y=x^3-x^2-9x+9 is:

$y=(x-1)(x-3)(x+3)$